Problem: In January, the average temperature $t$ hours after midnight in Mumbai, India, is given by $ T(t) = 24.5 -5.5 \sin \left(\dfrac{2\pi (t+1)}{24}\right)$. What is the coldest time of day in Mumbai? Give an exact answer.
The coldest time is when $\sin \left(\dfrac{2\pi (t+1)}{24}\right)$ is at its maximum. $\sin u$ is largest when $u$ is $\dfrac{\pi}{2}$ plus a multiple of $2\pi$. So the coldest time of day is when $\dfrac{2\pi (t+1)}{24} = \dfrac{\pi}{2} + 2\pi n$ for $n$, an integer. We can solve that equation for $t$ : $\begin{aligned}t + 1 &= 6 + 24 n \\ t &= 5 + 24 n. \end{aligned}$ Since changing the value of $n$ just adds a multiple of $24$ hours, it just changes what day it is: in each day, it's coldest $5$ hours after midnight. It's coldest $5$ hours after midnight each day, at $5$ : $00\text{ a.m.}$